Why Read This Book? - Index of

7.1 Bounded and Monotone Functions 209
EXERCISE 7.1.5 Show f(x) = 1/(x 2 + 1) is strictly decreasing on [0, ∞).
In Exercise 3.5.7, you proved the following theorem, even though we did not
have the language of monotonicity at the time.
Theorem 7.1.6 Let n be any positive integer. Then:
1. f(x) = x2n−1 is strictly increasing on the set of real numbers.
2. f(x) = x2n is strictly decreasing on (−∞, 0] and strictly increasing on [0, ∞).
In your pre-calculus work, you probably learned about the horizontal line
test, an intuitive way of determining whether a function f is one-to-one. If every
horizontal line in the xy-plane crosses the graph of f no more than once, then f is
one-to-one. If f does not map onto the real numbers, we can crop the codomain of
f down to Rng f so that f : S → Rng f is onto. Thus if f : S → Rng f is one-toone,
then f −1 : Rng f → S will exist. To find the expression for f −1 , you switched
the roles of x and y in the expression for f , then solved for y. You also learned
that the graph of f −1 could be sketched by reflecting the graph of f about the
diagonal line y = x. This reflection is the visual result of swapping x and y. Think
intuitively for a moment. If f is a strictly increasing function, then does it pass
the horizontal line test? If so, then f −1 exists on Rng f . Given this, can you say
anything about the monotonicity of f −1 ?
Theorem 7.1.7 If a function is strictly monotone on A, then it is one-to-one
on A.
EXERCISE 7.1.8 Prove Theorem 7.1.7 for the case of an increasing function.
(A similar argument would work for a decreasing function.)
As an immediate consequence of Exercise 7.1.8 and Theorem 4.4.7, we have
the following.
Corollary 7.1.9 If f is strictly monotone on A ⊆ S, then there exists an inverse
function f −1 : f(A) → A.
Letting A = S in Corollary 7.1.9, we have that if f : S → R is strictly monotone,
then f −1 : Rng f → S exists. Furthermore, the following is true.
Theorem 7.1.10 If f : S → R is strictly increasing (or strictly decreasing), then
f −1 : Rng(f) → S is strictly increasing (or strictly decreasing).
EXERCISE 7.1.11 Prove the case of Theorem 7.1.10 where f is increasing.
(A similar argument would work if f is decreasing.)
Theorems 7.1.6 and 7.1.10 imply that 2n√ x is strictly increasing on [0, ∞) and
2n−1 √ x is strictly increasing on R. Also, with the next exercise, the function f(x) = x r